# Discontinuous function $U$ with continuous preferences can be written as a composition of discontinuous & monotone function and a continuous function

Conjecture: Every discontinuous utility function $$U$$ representing continuous preferences can be written as $$U = f \circ g$$ for some continuous $$g$$ and discontinuous strictly monotone $$f$$.

The goal is to prove or disprove this conjecture.

We know that continuous utility implies continuous preferences, and running it through a strict monotone gives us the same preferences but with a discontinuous utility function. The conjecture is inspired from this.

• I recommend that you delete your counter-example and the edit from the body of the Q. Apr 7, 2022 at 14:12
• Are the continuous preferences themselves monotone? Apr 7, 2022 at 14:14
• @Giskard The preferences need not be monotone. (I have edited the post as well.) Apr 7, 2022 at 14:16
• Any assumptions on the domain of $U$? Apr 7, 2022 at 14:36
• @MichaelGreinecker The domain $X \subseteq \mathbb{R}^n$. But I would like to see these two cases, if possible: (1) $X \subseteq \mathbb{R}^n$ and (2) $X = \mathbb{R}^n$. Apr 7, 2022 at 14:57

I took the question as asking for a representation with $$f:\mathbb{R}\to\mathbb{R}$$. If we only require $$f$$ to be defined on a subset of $$\mathbb{R}$$, Amit's answer solves the problem.

Here is a proof for the case that the domain is connected (and second countable): Under these assumptions, there exists a continuous utility representation $$V:X\to\mathbb{R}$$. Since continuous functions map connected sets to connected sets, the image $$V(X)$$ is a, possibly unbounded, interval. I do the special case that $$V(X)=[a,b)$$ for some real numbers $$a$$ and $$b$$, the other cases can be dealt with by analogous methods. Now define the surjective continuous function $$g:X\to [a,\infty)$$ by $$g(x)=\frac{V(x)-a}{b-V(x)}.$$ Note that $$g$$ represents the preference ordering. Let $$r^*=U(x)$$ for any (and hence all) $$x\in g^{-1}(a).$$

Now define $$f:\mathbb{R}\to\mathbb{R}$$ so that for $$r\geq a$$ we have $$f(r)=U(x)$$ for any (and hence all) $$x\in g^{-1}(r)$$ and for $$r we have $$f(r)=r^*-|r-a|$$. It is easy to verify that $$f$$ is strictly increasing and $$U=f\circ g$$.

• Comments are not for extended discussion; this conversation has been moved to chat.
– 1muflon1
Apr 10, 2022 at 12:36

If the preference is continuous, reflexive, transitive and complete, then there exists a continuous utility representation $$g$$, and since $$U$$ also represents the same preference, so there must be a strictly increasing function $$f$$ such that $$U= f\circ g$$.

• Can you please elaborate the second part where you say that such an $f$ must exist? It may be true, but I can't convince myself that it really is the case. A proof would be appreciated! Apr 7, 2022 at 16:30
• Take it as an exercise. Try to construct such a $f$, it is easy.
– Amit
Apr 7, 2022 at 17:03
• The existence of $g$ comes from Debreu's theorem. Then define $f$ such that $f(g(x)) = U(x)$. Since $g$ and $U$ denote the same preferences, it follows that $f(g(x)) > f(g(y)) \iff U(x) \geq U(y) \iff g(x) \iff g(y)$ which further implies that $f$ is monotone. Does this work? Apr 8, 2022 at 0:18
• $f$ defined as $f(g(x)) = U(x)$ works provided you want $f:\text{Range}(g)\rightarrow\mathbb{R}$. Here $\text{Range}(g)$ will be an interval because $g$ is continuous. If you want $f:\mathbb{R}\rightarrow\mathbb{R}$, then you just need to complete the construction of $f$ outside the range of $g$.
– Amit
Apr 8, 2022 at 2:09
• @Pocket Cat If you want $f:\mathbb{R}\rightarrow\mathbb{R}$, see Michael's answer.
– Amit
Apr 8, 2022 at 7:30